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(1) For all agents i_{1}, i_{2}, … , i_{m} N, K_{i1}K_{i2} … K_{im}(A)Hence, K^{*}_{N}(A) iff (1) is the case for each m 1.
Proof.
Note first that
( 2 ) |
^{i 1 N} |
K_{ i 1} ( |
^{i 2 N} |
K_{ i 2} ( . . . ( |
^{i m - 1 N} |
K_{ i m - 1} ( |
^{i m N} |
K_{ i m}( A ) ) ) ) ) |
= |
^{i 1 N} |
K_{ i 1} ( |
^{i 2 N} |
K_{ i 2} ( . . . ( |
^{i m - 1 N} |
K_{ i m - 1} (K^{ 1}_{N} ( A ) ) ) ) ) |
= |
^{i 1 N} |
K_{ i 1} ( |
^{i 2 N} |
K_{ i 2} . . . ( |
^{i m - 2 N} |
K_{ i m - 2} ( K^{ 2}_{N} ( A ) ) ) ) |
= . . . |
= |
^{i 1 N} |
K_{ i 1}( K^{ m - 1}_{N} ( A ) ) |
= | K^{ m}_{N} ( A ) |
By ( 2 ), K^{ m}_{N} ( A ) K_{ i 1}K_{ i 2} . . . K_{ i m}( A ) for i_{ 1}, i_{ 2}, . . ., i_{ m} N so if K^{ m}_{N} ( A ) then condition ( 1 ) is satisfied. Condition ( 1 ) is equivalent to
^{i 1 N}K_{ i 1} (
^{i 2 N}K_{ i 2} ( . . . (
^{i m - 1 N}K_{ i m - 1} (
^{i m N}K_{ i m} ( A ) ) ) ) )
so by ( 2 ), if ( 1 ) is satisfied then K^{ m}_{N} ( A ).
First published: August 27, 2001
Content last modified: August 27, 2001